The essence of binary search is not "monotonicity", but "boundary". As long as a certain property is found that divides the entire interval into two, the boundary point can be found using binary search.
boolean check(int x) {
}
int search(int left, int right) {
while (left < right) {
int mid = (left + right) >> 1;
if (check(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
boolean check(int x) {
}
int search(int left, int right) {
while (left < right) {
int mid = (left + right + 1) >> 1;
if (check(mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}
When doing binary search problems, you can follow the following routine:
Note that the advantage of these two templates is that they always keep the answer within the binary search interval, and the value corresponding to the end condition of the binary search is exactly at the position of the answer. For the case that may have no solution, just check whether the $left$ or $right$ after the binary search ends satisfies the problem.
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